Metamath Proof Explorer


Theorem absdifltd

Description: The absolute value of a difference and 'less than' relation. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses absltd.1 ( 𝜑𝐴 ∈ ℝ )
absltd.2 ( 𝜑𝐵 ∈ ℝ )
absltd.3 ( 𝜑𝐶 ∈ ℝ )
Assertion absdifltd ( 𝜑 → ( ( abs ‘ ( 𝐴𝐵 ) ) < 𝐶 ↔ ( ( 𝐵𝐶 ) < 𝐴𝐴 < ( 𝐵 + 𝐶 ) ) ) )

Proof

Step Hyp Ref Expression
1 absltd.1 ( 𝜑𝐴 ∈ ℝ )
2 absltd.2 ( 𝜑𝐵 ∈ ℝ )
3 absltd.3 ( 𝜑𝐶 ∈ ℝ )
4 absdiflt ( ( 𝐴 ∈ ℝ ∧ 𝐵 ∈ ℝ ∧ 𝐶 ∈ ℝ ) → ( ( abs ‘ ( 𝐴𝐵 ) ) < 𝐶 ↔ ( ( 𝐵𝐶 ) < 𝐴𝐴 < ( 𝐵 + 𝐶 ) ) ) )
5 1 2 3 4 syl3anc ( 𝜑 → ( ( abs ‘ ( 𝐴𝐵 ) ) < 𝐶 ↔ ( ( 𝐵𝐶 ) < 𝐴𝐴 < ( 𝐵 + 𝐶 ) ) ) )